## Elementary Geometry for College Students (6th Edition)

$33 \sqrt3$
We first find the height of the trapezoid. It is a leg of the 30-60-90 triangle containing the side of length 6, so we obtain: $h= 6 \times \frac{\sqrt3}{2} = 3\sqrt3$ We also find the lengths of AE and FD, which are equal: $AE=FD = 6 \times \frac{1}{2} = 3$ Thus: $A = 1/2(b_1+b_2)(h) = 1/2(8+14)(3\sqrt3)=33\sqrt3$